Question

$P$ and $Q$ are the mid-points of the opposite sides $AB$ and $CD$ of a parallelogram $ABCD.AQ$intersects $DP$ at $S$ and $BQ$ intersects $CP$ at $R$. Show that $PRQS$ is a parallelogram.

Answer 1

Proof.

$Q$ is the midpoint of $AB$ and $P$ is the midpoint of $CD$.

Now,

$AB\left|\right|CD$, $AP\left|\right|QC$

And,

$$\begin{gathered} AB=DC \\ \Rightarrow \frac{1}{2}AB=\frac{1}{2}DC \\ \therefore AP=QC \end{gathered}$$

Also

$AP\left|\right|QC$ and $AP=QC$

$APCQ$ is a parallelogram.

$AQ\left|\right|PC$ or $SQ\left|\right|PR$

Again,

$AB\left|\right|DC$means $\frac{1}{2}AB=\frac{1}{2}DC$

So, $BP=QD$

Now, $BP\left|\right|QD$ and $BP=QD$

$BPDQ$ is a parallelogram

So, $PD\left|\right|BQ$or $PS\left|\right|QR$

Thus, $SQ\left|\right|RP$ and $PS\left|\right|QR$

Hence, $PRQS$ is a parallelogram.